EXPLANATION OF OXYGEN IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 7, 2015 Oxygen is a chemical element with symbol O and atomic number 8. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Oxygen including the following ground state electron configuration: 1s2.2s2.2px2.2py1.2pz1. According to the “Ionization energies of the elements-WIKIPEDIA” we observe ( in eV ) the following energies: E1 = 13.618 , E2 = 35.117 , E3 = 54.9355 , E4 = 77.41355 , E5 = 113.899 , E6 = 138.1197 . E7 = 739.29 , and E8 = 871.41 . ( See also my papers in my FUNDAMENTAL PHYSICS CONCEPTS). For simplicity we start with the E4 and E3 , while the E2 and E1 are explained in the second chapter. EXPLANATION OF E4 = 77.41355 eV AND E3 = 54.9355 eV The 2px2 , 2py1 , and 2pz1 of charge (- 4e) under a perfect screening of the spherical shells 1s2 and 2s2 orbitals should lead to the effective ζ = 4, because +8e -2e -2e = +4e. However the deformation of shells leads to ζ > 4 . Of course in the absence of the electrons 2px2 and 2py1 one gets the E(2pz1) = - E4 which is the binding energy of the one electron with n = 2 based on the Bohr model. In this case of n = 2 we must find the effective ζ1 > 4, while for a perfect screening (ζ = 4) due to the spherical shells of 1s2 and 2s2 we should find E (2pz1) = (-13.6) ζ2/22 = (-13.6)42/22 = - 54.4 eV However the 2pz1 penetrates the 2s2 which leads to the deformations not only of spherical shells 1s2 and 2s2 but also of 2pz1. Thus writing E4 = 77.41355 eV = - E(2pz1) = - (- 13.6) ζ12/22 we get ζ1 = 4.77 > 4 . Then adding the second electron 2py1 one gets the binding energy E (2pz1 + 2py1) = -( E4 + E3 ). That is, E(2pz1 + 2py1) = 2(-13.6)ζ22/22 = - (E4 + E3 ) = - (77.41355 + 54.9355) = - 132.349 eV Therefore one gets ζ2 = 4.41 > 4 Because of parallel spin (S = 1 ) here we observe two separated electrons of 2pz1 and 2py1 with mutual electric and magnetic repulsions: Fem = Fe + Fm In other words the two electrons of 2pz1 and 2py1 try to be far away out of the shells 1s2 and 2s2. ' ' EXPLANATION OF E2 = 35.117 eV BASED ON THE BOHR FORMULA AND OF E1 = 13.618 eV BASED ON MY FORMULA OF 2008 Adding the third electron 2px1 one gets the binding energy of the three separated electrons which circulate outsid08'e the shells 1s2 and 2s2 : E( 2px1 + 2py1 + 2pz1) = 3(-13.6)ζ32/22 = - (E2 + E3 + E4 ) = - 167.466 Therefore we get ζ3 = 4 Here ζ3 = 4 means that the three electrons of 2px1, 2py1, and 2pz1 orbitals with S =1 giving electric and magnetic repulsions try to be at symmetrical positions leading to a perfect screening of the spherical shells 1s2 and 2s2 . However in the absence of 1px1 the two electrons of 2py1 and 2pz1 break the symmetry and lead to ζ2 = 4.41 > 4 . Moreover in the absence of 2px1 and 2py1 the one electron of 2pz1 orbital breaks more the symmetry and leads to ζ1 = 4.77 > 4. That is, ζ1 > ζ2 > ζ3 . Now adding the fourth electron one sees that it is the first electron being removed. It is one of the 2px2 pair, because of a vibration energy (Ev) discovered in my paper of 2008. Since all theories after the Bohr model could not provide any satisfactory formula for the explanation of the two-electron atoms, I published in 2008 the paper according to which two electrons of opposite spin operate with a vibration energy: Ev = (16.95ζ - 4.1)/n2 where ζ is the effective Zeff under a screening and n is the quantum number. Of course, in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of the 2px2 pair, today many physicists believe incorrectly that it is due to the Coulomb repulsion between the two electrons. Under such fallacious ideas I published my paper of 2008. Here the binding energy of the two electrons of opposite spin E(2px2) = -( E1 + E2) is given by E(px2 ) = + (16.95)ζ - 4.1 / 22 = - ( E1 + E2) Thus the total binding energy of the four electrons can be written as E( 2px2 + 2py1 + 2pz1) = + (16.95)ζ - 4.1 /22 + 2(-13.6)ζ2/22 = - ( Ε1 + Ε2 + Ε3 + Ε4) Since - ( Ε1 + Ε2 + Ε3 + Ε4 ) = - 181 eV we may write 13.6ζ2 - 4.2375ζ -179.975 = 0 Then solving for ζ we get ζ = 3.8 < 4. Since ζ < 4 cannot exist we suggest that the correct ζ = 4 of a perfect screening gives n > 2. Under this condition we find that n = 2.11 . It means that the orbitals 2px2 , 2py1 , and 2pz1 do not lead to the deformation of 1s2 and 2s2 but differ from the symmetry of the (2px1 + 2py1+ 2pz1 ) , because here the paired electrons of opposite spin do not repel magnetically the 2py1 and 2pz1 . Thus, under the reduction of the magnetic repulsions between 2px2 and 2py1 and 2pz1 we observe the same perfect screening under n = 2.11 . ' ''' '''EXPLANATION OF E5 = 113.899 eV AND E6 = 138.1197 eV For simplicity we start with the E6 = -E( 2s1). According to the quantum mechanics the one electron of the orbital 2s1 penetrates the 1s2 shell. Thus it leads to the deformations of both 1s2 and 2s1 spherical shells giving an effective ζ > 6 because the charge (- 2e) of the two electrons of 1s2 screens the charge (+8e) of nucleus. Since n = 2 we may write E5 = 138.1197 eV = - E(2s1) = - (-13.6)ζ2/22 Therefore one gets ζ2 = 40.62 and ζ = 6.37 > 6 Then adding the second electron of opposite spin we get the binding energy, E(2s2), of the two electrons of opposite spin for n = 2 given by my formula of 2008: E(2s2) = + (16.95)ζ - 4.1 / 22 = - (E5 + E6 ) Or 6.8ζ2 - 4.2375ζ + 1.025 - (E5 + E6) = 0 Since - ( E5 + E6 ) = -252.08 we may write 6.8ζ2 - 4.2375ζ - 251 = 0 Thus, solving for ζ we get ζ = 6.39 > 6 In other words we observe that the repulsions 1s2 - 2s1 and 1s2 - 2s2 give almost the same effective ζ because the two electrons (2s2) of opposite spin behave like one particle. Whereas in the case of the three electrons of S = 1 we observe a perfect screening, because the three electrons of 2px1, 2py1, and 2pz1 , interact from symmetrical positions. ' ' EXPLANATION OF Ε7 = 739.29 eV AND E8 = 871.41 eV As in the case of the helium the ionization energy E8 = 871.41 eV = - E(1s1) is due to the one remaining electron of 1s1 with n = 1 given by applying the simple Bohr model for Z = 8 as E8 = - (-13.6057)Z2 /12 = (13.6057)82/12 = 870.7648 eV. Surprisingly here we see that the value of 870.7648 eV given by the Bohr model is smaller than the experimental value of 871.41 eV, because after the ionizations the nuclear charge becomes much greater than the charge of the one electron. Under this condition I suggest that the quantum number n = 1 becomes n < 1. Therefore the above equation could be written as E8 = 871.41 eV = - (-13.6057)Z2 /n2 = (13.6057)82/n2 Then solving for n we get n =0.9996297 In the same way for calculating the E7 = 739.29 eV we must write my formula of 2008 as E7 = 739.29 eV = - 871.41 - E(1s2) = - 871.41 - [ (-27.21)82 + (16.95)8 - 4.1] / n2 Then solving for n we get n = 0.999764 Here 0.999764 > 0.9996297 because the second electron increases the electron charge with respect to the nuclear charge. Category:Fundamental physics concepts